A light rod of length 1 m is pivoted at its centre, with two masses...

Let the length of the rod be $L = 1\text{ m}$. Since the rod is pivoted at its centre, the distance of each mass from the pivot is $r = \dfrac{L}{2} = 0.5\text{ m}$.

The forces acting on the system are the weights of the two masses. The net torque $\tau$ about the pivot is the difference between the torques produced by the $5\text{ kg}$ and $2\text{ kg}$ masses:

$\tau = m_1 g r - m_2 g r = (m_1 - m_2) g r$

Substituting the given values (taking $g = 9.8\text{ m/s}^2$):

$\tau = (5 - 2) \times 9.8 \times 0.5 = 3 \times 4.9 = 14.7\text{ N m}$

The total moment of inertia $I$ of the system about the pivot, assuming the rod is light (massless), is the sum of the moments of inertia of the two point masses:

$I = m_1 r^2 + m_2 r^2 = (m_1 + m_2) r^2$

$I = (5 + 2) \times (0.5)^2 = 7 \times 0.25 = 1.75\text{ kg m}^2$

Using Newton's second law for rotation, $\tau = I \alpha$, we can find the initial angular acceleration $\alpha$:

$\alpha = \dfrac{\tau}{I} = \dfrac{14.7}{1.75} = 8.4\text{ rad/s}^2$